# proof of existence of the essential supremum

Suppose that $(\Omega,\mathcal{F},\mu)$ is a $\sigma$-finite measure space and $\mathcal{S}$ is a collection of measurable functions $f\colon\Omega\rightarrow\mathbb{\bar{R}}$. We show that the essential supremum of $\mathcal{S}$ exists and furthermore, if it is nonempty then there is a sequence $f_{n}\in\mathcal{S}$ such that

 ${\mathrm{ess}\sup}\,\mathcal{S}=\sup_{n}f_{n}.$

As any $\sigma$-finite measure is equivalent to a probability measure (http://planetmath.org/AnySigmaFiniteMeasureIsEquivalentToAProbabilityMeasure), we may suppose without loss of generality that $\mu$ is a probability measure. Also, without loss of generality, suppose that $\mathcal{S}$ is nonempty, and let $\mathcal{S}^{\prime}$ consist of the collection of maximums of finite sequences of functions in $\mathcal{S}$. Then choose any continuous and strictly increasing $\theta\colon\mathbb{\bar{R}}\rightarrow\mathbb{R}$. For example, we can take

 $\theta(x)=\left\{\begin{array}[]{ll}x/(1+|x|),&\textrm{if }|x|<\infty,\\ 1,&\textrm{if }x=\infty,\\ -1,&\textrm{if }x=-\infty.\end{array}\right.$

As $\theta(f)$ is a bounded and measurable function for all $f\in\mathcal{S}^{\prime}$, we can set

 $\alpha=\sup\left\{\int\theta(f)\,d\mu:f\in\mathcal{S}^{\prime}\right\}.$

Then choose a sequence $g_{n}$ in $\mathcal{S}^{\prime}$ such that $\int\theta(g_{n})\,d\mu\rightarrow\alpha$. By replacing $g_{n}$ by the maximum of $g_{1},\ldots,g_{n}$ if necessary, we may assume that $g_{n+1}\geq g_{n}$ for each $n$. Set

 $f=\sup_{n}g_{n}.$

Also, every $g_{n}$ is the maximum of a finite sequence of functions $g_{n,1},\ldots,g_{n,{m_{n}}}$ in $\mathcal{S}$. Therefore, there exists a sequence $f_{n}\in\mathcal{S}$ such that

 $\{f_{1},f_{2},\ldots\}=\{g_{n,m}:n\geq 1,1\leq m\leq m_{n}\}.$

Then,

 $f=\sup_{n}f_{n}.$

It only remains to be shown that $f$ is indeed the essential supremum of $\mathcal{S}$. First, by continuity of $\theta$ and the dominated convergence theorem,

 $\int\theta(f)\,d\mu=\lim_{n\rightarrow\infty}\int\theta(g_{n})\,d\mu=\alpha.$

Similarly, for any $g\in\mathcal{S}$,

 $\int\theta(f\vee g)=\lim_{n\rightarrow\infty}\int\theta(g_{n}\vee g)\,d\mu\leq\alpha.$

It follows that $\theta(f\vee g)-\theta(f)$ is a nonnegative function with nonpositive integral, and so is equal to zero $\mu$-almost everywhere. As $\theta$ is strictly increasing, $f\vee g=f$ and therefore $f\geq g$ $\mu$-almost everywhere.

Finally, suppose that $g\colon\Omega\rightarrow\mathbb{\bar{R}}$ satisfies $g\geq h$ ($\mu$-a.e.) for all $h\in\mathcal{S}$. Then, $g\geq f_{n}$ and,

 $g\geq\sup_{n}f_{n}=f$

$\mu$-a.e., as required.

Title proof of existence of the essential supremum ProofOfExistenceOfTheEssentialSupremum 2013-03-22 18:39:25 2013-03-22 18:39:25 gel (22282) gel (22282) 7 gel (22282) Proof msc 28A20